I’m using 500VA transformer. Theoretically, on 50V max current is about 10A. Since both 2N3055 can support up to 30A (15A each), the limit depend on the transformer it self. If you have larger transformer maybe u can increase this value. But you need a bigger heat sink. I never try to test the current limit before (afraid of burning it).

Some questions & comments on this PSU:1. Your stripboard layout does a discrete diode bridge rectifier, while the bill of materials calls for a full wave rectifier.2. The LM384 is rated 3v to 38v, yet as I see it this circuit supplies it with >50V DC (unregulated), or am I missing something?Thanks

answer #1ops! I plan to use seperate diode but when I found 1 old full wave rectifier from my spare part stock I forgot about it. Sorry for that. You can see at the picture, there are no diode on it. I attach the full wave rectifier near the transformer output.

answer #2I didn't know about the limit actually (38V) but what I know that I did measure the transformer output and the reading show me 50V. Anybody have some explaination about this?

I found your power supply to be just the thing for a home anodizing setup because of the constant current mode. Could you give a few more details on the operation. I have all the parts except the meters and zener diode. I'll probably just plug in a digital meter instead of panel meters . Is there a range of values that will work for the zener diode I cant seem to find a 9.1 in my junk box . would a 12v or a 13v work?

Thanks for your reply. My experience in electrical is very basic, mostly self taught from internet research. I can build from your plans but I dont have the knowledge to substitute parts or modify a circut.After a lot of searching your power supply is the best all around I've found for my purpose (anodize and general shop use).

This is an ambitious project. To answer your question about the 50VRMS measurement you made: RMS is a measure of how many volts DC would deliver the same power to some load - so if you connect a 50 Ohm resistor directly across that transformer's output, it will sink about 50/50 = 1A of current, and dissipate 1A * 50V = 50W of power.

But the transformer's output is AC, and this means the AVERAGE power will be 50W. In fact, the PEAK power dissipated in the 50 Ohm load (which occurs 50 times per second) is much higher. The peak voltage of a sinusiodal source is about 1.4 times higher than the RMS voltage!

So if you measure the voltage across your 2200uF reservoir capacitor AND the transformer is REALLY supplying 50V RMS, then you'll find about 70V or so. If you find 50V there, then your transformer is providing 50 / 1.4 = 36V RMS.

It's better to make these measurements with a small load because, under no load, transformer outputs are likely to exceed their rated voltages.

I seriously doubt that your transformer is really 50V RMS, because that poor little LM324 would certainly pop when supplied with 70V DC. What DC voltage do you measure across that 2200uF cap?

The 2200uF reservoir capacitor is quite small. By my calculations, under a 1A load, it will discharge roughly 5V between each 100Hz rectified cycle.

Under a 2A load, this will be more like 9V.

Keep this in mind - heavy loads may cause your DC power rail to drop too low. The the solution is to use a larger reservoir capacitor.

That little BC109C transistor is in trouble - The 2N3055 has a tiny hfe of 30 or so, meaning that the BC109C must provide up to 3% of the total current load! If the PSU output is set low (say 10V) then this little fella must drop nearly 40V! (Assuming the unregulated power rail is 50V). If he's also sourcing 3% of a 1A load to the bases of the 2N3055s, he is doomed, because he's dissipating 40V x 30mA = 1.2W. If memory serves, he's good for 300mW only.

Worst case calculations would assume that the PSU should be able to supply near-zero Volts ouput, and almost ALL of this will be dropped across the BC109C! Now, knowing the maximum current that this PSU is designed to source, remember that 30% of this is being sourced by the BC109C. For instance, if you plan on a 3A limit, down to 3V mimumum output, then your calculations might be like this:

(50V - 3V) x 3A * 3% = 4.2W

That's a lot of power for a small transistor, and this is one reason why lab power supplies are so expensive! A cheap solution to this dilemma is to use a 50W power transistor (not a BC109C) mounted on a heat sink - I think the LM324 would be able to drive it. Better still, use a power darlington, to keep the op-amp as unloaded as possible.

One more observation: With the two 2N3055 emitters connected together like that, only one of the two transistors will be doing any work. This is because the transistor with the largest Vbe will not be biased as much into into conduction as the other, which will thus be lumped with almost all of the emitter current. You might as well use just one 2N3055.

For such load sharing to work, you need to buffer the emitters with a small resistance, say 1 Ohm. This gives the emitter voltage some freedom to move, and imbalances between the transitors are less important.

The buffer resistors will drop some voltage, of course. With a 4A load, each 2N3055 transistor is sourcing 2A, and their emitter resistors (1 Ohm each) will thus each drop 2V. That's 2V * 2A = 4W of power each. So make them sturdy.

>After in depth study of lab power >supply Schematic,I suggest some >modifications.Use a taped >transformer(17+17V)to control power >Dissipation of PASS TRS 2N3055. >Use a Voltage sensing circuit to >select 17V AC or 36V AC.Replace >BC109 with BC160 or BD135.Add 0.1 >Ohm 3W Resistors in between >2N3055,Emitter & out put. To save >LM234 use spare power supply of >24V.I think this will help.

Hi, Nice device.Quez: I've got another ampere meter, not 1kohm. Can I juse it if I switch the 560ohm resistant to another one so the resistor + amperemeter = 1560 ohm ? I belive my meter is 700 ohm, so then I nees a 860 ohm resistor. Thanx

One more problem to be solved before it can be called "Lab Power Supply". The output capacitor of 2200 should not be that big. It will supply excessive current in the moment of load connection, and before of the current limitting. If the 100uF is not enough, then something is wrong with the overall stability.